­If xy = 3(x + 1) + y; and, x and y are integers, x could be any of

 

Ah yes, Bunuel ! The question would become logical then.­

xy - 3x - y = 3

xy - 3x - y + 3 = 6

(x - 1)(y - 3) = 6

(1, 6) (-1, -6) (6, 1) (-6, -1) (2, 3) (3, 2) (-2, -3) (-3, -2)

x - 1 = 1, -1, 6, -6, 2, 3, -2, -3

x = 2, 0, 7, -5, 3, 4, -1, -2­

What's the best way to solve a problem like this?

 

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You can use the method shown here. Or do the following:

There are several ways to solve this question through algebraic manipulations. You can use the method shown here. Or do the following:

Essentially, our goal is to arrive at an equation of the form \(x = expression \ with \ y \ only\) or \(y = expression \ with \ x \ only\). 

Next, let's attempt to split the fraction to isolate \(x\) in a single term:

Since \(y\) is an integer, for \(3+ \frac{6}{x-1}\) to be an integer, \(\frac{6}{x-1}\) must also be an integer. This means that \(x-1\) must be a divisor of 6. From the options, only if x = 5, x - 1 is not a factor of 6, making it the correct answer.

Answer: B.